What you describe here is a well known sequencing of vertex layers while running through an nD (hyper)cube orthogonally to its (hyper)body diagonal. In fact that sequence of intersections are being described by

- o3o3o3o3...3o3o (top vertex)
- q3o3o3o3...3o3o (simplex, i.e. vertex figure)
- o3q3o3o3...3o3o
- o3o3q3o3...3o3o
- o3o3o3q3...3o3o
- ...
- o3o3o3o3...3q3o
- o3o3o3o3...3o3q (dual simplex, i.e. vertex figure)
- o3o3o3o3...3o3o (bottom vertex)

Here I used the unit-length sided (hyper)cube, i.e. the edge lengths of these sections are the diagonals of a (unit-sided) square, i.e. q=sqrt(2).

Note that the CD diagram of an nD (hyper)cube clearly has n nodes, accordingly all these sections will have n-1 nodes only.

The remainder then is just to splitt n into even and odd values, for if n would be even, then n-1 becomes odd and the above sequence will have a medial entry (which indeed for n=4, the starting figure being a tesseract, happens to be o3q3o, i.e. describes an octahedron).

On the other hand, if n would be odd, then n-1 becomes even and in the above sequence the "medial section" would run midwise between the vertex layers o3o3..3o3q3o3o3...3o3o and o3o3..3o3o3q3o3...3o3o (both ends of the CD having same size here) and the thus obtained midwise cross-section of the nD (hyper)cube becomes o3o3...3o3y3y3o3...3o3o, i.e. (n-1)/2-truncated simplex, where y represents the medial section of half-square triangles, i.e. y=sqrt(2)/2. E.g. for n=3 we'd get (3-1)/2=1, i.e. the truncated triangle y3y (because of (3-1)=2: 2D simplex), which is nothing but the hexagon you mentioned.

--- rk